91影视

• $F(x,y)=\frac{xe\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\mathbf{i}+\frac{ye\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\mathbf{j}$
• moving an object around the unit circle
• starting and ending at $(1,0)$.

The work done by a vector field F moving a particle along the curve C is computed by the following line integral:

${鈭�}_C\mathbf{F}路d\mathbf{r}$

Hence, evaluate the line integral ${鈭�}_C\mathbf{F}路d\mathbf{r}$to evaluate the required work done.

Use Green's Theorem to evaluate the required the line integral ${鈭�}_C\mathbf{F}路d\mathbf{r}$.

Green's Theorem states that,

"Let R be a region in the plane with smooth boundary curve C oriented counterclockwise by $\mathbf{r}\left(t\right)=鉄�\left(x\right(t),y(t\left)\right)鉄�$for $a鈮�t鈮�b$.

If a vector field is defined on R, then,

${鈭�}_C\mathbf{F}路d\mathbf{r}={鈭�}_R\left(\frac{鈭�F_2}{鈭�x}-\frac{鈭�F_1}{鈭�y}\right)dA."鈥�鈥�\left(1\right)$

For the vector field $\mathbf{F}(x,y)=\frac{x{e}^{\sqrt{x^2+y^2}}}{\sqrt{x^2+y^2}}\mathbf{i}+\frac{y{e}^{\sqrt{x^2+y^2}}}{\sqrt{x^2+y^2}}\mathbf{j}$,

$F_1(x,y)=\frac{x{e}^{\sqrt{x^2+y^2}}}{\sqrt{x^2+y^2}}$and $F_2(x,y)=\frac{y{e}^{\sqrt{x^2+y^2}}}{\sqrt{x^2+y^2}}$.

Now, first find $\frac{鈭�F_2}{鈭�x}$and $\frac{鈭�F_1}{鈭�y}$.

Then,

$$\begin{gathered} \frac{鈭�F_2}{鈭�x}=\frac{鈭�}{鈭�x}\left(\frac{y{e}^{\sqrt{x^2+y^2}}}{\sqrt{x^2+y^2}}\right) \\ =\frac{-xy\left(1-\sqrt{x^2+y^2}\right)e^{\sqrt{x^2+y^2}}}{{\left(x^2+y^2\right)}^{3/2}} \end{gathered}$$

and,

Now, use Green's Theorem (1) to evaluate the integral ${鈭�}_C\mathbf{F}路d\mathbf{r}$as follows:$$\begin{gathered} {鈭�}_C\mathbf{F}路d\mathbf{r}={鈭�}_R\left(\frac{鈭�F_2}{鈭�x}-\frac{鈭�F_1}{鈭�y}\right)dA \\ ={鈭�}_R\left(\frac{-xy\left(1-\sqrt{x^2+y^2}\right)e^{\sqrt{x^2+y^2}}}{{\left(x^2+y^2\right)}^{3/2}}-\frac{-xy\left(1-\sqrt{x^2+y^2}\right)e^{\sqrt{x^2+y^2}}}{{\left(x^2+y^2\right)}^{3/2}}\right)dA \\ ={鈭�}_{R}0dA \\ =0. \end{gathered}$$

Therefore, the required work done is 0.